SOLUTION: How do you find the solutions of the equation sin2x + cos(x)=0

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Question 172849This question is from textbook
: How do you find the solutions of the equation sin2x + cos(x)=0 This question is from textbook

Found 2 solutions by solver91311, Alan3354:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Your problem statement is ambiguous. Do you mean sin%5E2%28x%29%2Bcos%28x%29=0 or sin%282x%29%2Bcos%28x%29=0?

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sin2x + cos(x)=0
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sin^2 = 1-cos^2
1-cos^2 + cos = 0
cos^2 - cos - 1 = 0
A quadratic in cos(x)
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-1=5.

Discriminant d=5 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+5+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+5+%29%29%2F2%5C1+=+1.61803398874989
x%5B2%5D+=+%28-%28-1%29-sqrt%28+5+%29%29%2F2%5C1+=+-0.618033988749895

Quadratic expression 1x%5E2%2B-1x%2B-1 can be factored:
1x%5E2%2B-1x%2B-1+=+%28x-1.61803398874989%29%2A%28x--0.618033988749895%29
Again, the answer is: 1.61803398874989, -0.618033988749895. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-1+%29

cos(x) = 1/2 + sqrt(5)/2 =~1.618
cos>1 is not a real number, so ignore it.
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cos(x) = 1/2 - sqrt(5)/2 =~ -0.618
x = 128.173º